HOOK FORMULAS FOR SKEW SHAPES

The celebrated hook-length formula gives a product formula for the number of standard Young tableaux of a straight shape. In 2014, Naruse announced a more general formula for the number of standard Young tableaux of skew shapes as a positive sum over excited diagrams of products of hook-lengths. We give an algebraic and a combinatorial proof of Naruse’s formula, by using factorial Schur functions and a generalization of the Hillman-Grassl correspondence, respectively. Our main results are two q-analogues of Naruse’s formula for the skew Schur functions and for counting reverse plane partitions of skew shapes. We also apply our results to border strip shapes and their generalizations. In particular, we obtain new summation formulas for the number of alternating permutations in terms of certain Dyck paths, and their q-analogues.

1. Introduction 1.1.Foreword.The classical hook-length formula (HLF) for the number of standard Young tableaux (SYT) of a Young diagram, is a beautiful result in enumerative combinatorics that is both mysterious and extremely well studied.In a way it is a perfect formula -highly nontrivial, clean, concise and generalizing several others (binomial coefficients, Catalan numbers, etc.)The HLF was discovered by Frame, Robinson and Thrall [FRT] in 1954, and by now it has numerous proofs: probabilistic, bijective, inductive, analytic, geometric, etc. (see §11.2).Arguably, each of these proofs does not really explain the HLF on a deeper level, but rather tells a different story, leading to new generalizations and interesting connections to other areas.In this paper we prove a new generalization of the HLF for skew shapes which presented an unusual and interesting challenge; it has yet to be fully explained and understood.
For skew shapes, there is no product formula for the number f λ/µ of standard Young tableaux (cf.Section 10).Most recently, in the context of equivariant Schubert calculus, Naruse presented and outlined a proof in [Naru] of a remarkable generalization on the HLF, which we call the Naruse hook-length formula (NHLF).This formula (see below), writes f λ/µ as a sum of "hook products" over the excited diagrams, defined as certain generalizations of skew shapes.These excited diagrams were introduced by Ikeda and Naruse [IN1], and in a slightly different form independently by Kreiman [Kre1,Kre2] and Knutson, Miller and Yong [KMY].They are a combinatorial model for the terms appearing in the formula for Kostant polynomials discovered independently by Andersen-Jantzen-Soergel [AJS, Appendix D] and Billey [Bil] (see Remark 4.2 and §11.3).These diagrams are the main combinatorial objects in this paper and have difficult structure even in nice special cases (cf.§ 8.1).
The goals of this paper are threefold.First, we give Naruse-style hook formulas for the Schur function s λ/µ (1, q, q 2 , . ..), which is the generating function for semistandard Young tableaux (SSYT) of shape λ/µ, and for the generating function for reverse plane partitions (RPP) of the same shape.Both can be viewed as q-analogues of NHLF.In contrast with the case of straight shapes, here these two formulas are quite different.Even the summations are over different sets -in the case of RPP we sum over pleasant diagrams which we introduce.The proofs employ a combination of algebraic and bijective arguments, using the factorial Schur functions and the Hillman-Grassl correspondence, respectively.While the algebraic proof uses some powerful known results, the bijective proof is very involved and occupies much of the paper.
Second, as a biproduct of our proofs we give the first purely combinatorial (but non-bijective) proof of Naruse's formula.We also obtain trace generating functions for both SSYT and RPP of skew shape, simultaneously generalizing classical Stanley and Gansner formulas, and our q-analogues.We also investigate combinatorics of excited and pleasant diagrams and how they related to each other, which allow us simplify the RPP case.
Third, we apply our results to the case of border strips δ m+2 /δ m and more general thick strips δ m+2k /δ m .We obtain new summation formulas for two different q-Euler polynomials and a host of determinant formulas.
1.2.Hook formulas for straight and skew shapes.We assume here the reader is familiar with the basic definitions, which are postponed until the next two sections.
The standard Young tableaux (SYT) of straight and skew shapes are central objects in enumerative and algebraic combinatorics.The number f λ = | SYT(λ)| of standard Young tableaux of shape λ has the celebrated hook-length formula (HLF): Theorem 1.1 (HLF; Frame-Robinson-Thrall [FRT]).Let λ be a partition of n.We have: (1.1) , where h(u) = λ i − i + λ j − j + 1 is the hook-length of the square u = (i, j).
When µ = ∅, there is a unique excited diagram D = ∅, and we obtain the usual HLF.
This formula can be viewed as q-analogue of the HLF.In fact, one can derive HLF (1.1) from (1.3) by Stanley's theory of P -partitions [S3,Prop. 7.19.11]or by a geometric argument [Pak,Lemma 1].
Here we give the following natural analogue of NHLF (1.3).
By analogy with the straight shape, Theorem 1.4 implies NHLF, see Proposition 3.4.We prove Theorem 1.4 in Section 4 by using algebraic tools.
1.4.Hook formulas for reverse plane partitions via bijections.In the case of staight shapes, the enumeration of RPP can be obtained from SSYT, by subtracting (i − 1) from the entries in the i-th row.In other words, we have: (1.5) π∈RPP(λ) u) .
Note that the above relation does not hold for skew shapes, since entries on the i-th row of a skew SSYT do not have to be at least (i − 1).
Formula (1.5) has a classical combinatorial proof by the Hillman-Grassl correspondence [HiG], which gives a bijection Φ between RPP ranked by the size and nonnegative arrays of shape λ ranked by the hook weight.We view RPP of skew shape λ/µ as a special case of RPP of shape λ.The major technical result of the paper is Theorem 7.7, which states that the restriction of Φ gives a bijection between SSYT of shape λ/µ and arrays of nonnegative integers of shape λ with zeroes in the excited diagram and certain nonzero cells (excited arrays, see Definition 7.1).In other words, we fully characterize the preimage of the SSYT of shape λ/µ under the map Φ.This and the properties of Φ allows us to obtain a number of generalizations of Theorem 1.4 (see below).
The proof of Theorem 7.7 goes through several steps of interpretations using careful analysis of longest decreasing subsequences in these arrays and a detailed study of structure of the resulting tableaux under RSK.We built on top of the celebrated Greene's theorem and several Gansner's results.As a corollary of our proof of Theorem 7.7, we obtain the following generalization of formula (1.5).This result is natural from enumerative point of view, but is unusual in the literature (cf.Section 10 and §11.5), and is completely independent of Theorem 1.4.
The theorem employs a new family of combinatorial objects called pleasant diagrams.These diagrams can be defined as subsets of complements of excited diagrams (see Theorem 6.10), and are technically useful.This allows us to write the RHS of (1.6) completely in terms of excited diagrams (see Corollary 6.17).Note also that as corollary of Theorem 1.5, we obtain a combinatorial proof of NHLF (see §6.4).
1.5.Further extensions.One of the most celebrated formula in enumerative combinatorics is MacMahon's formula for enumeration of plane partitions, which can be viewed as a limit case of Stanley's trace formula (see [S1, S2]): Here tr(π) refers to the trace of the plane partition.
These results were further generalized by Gansner [G1] by using the properties of the Hillman-Grassl correspondence combined with that of the RSK correspondence (cf.[G2]).
For SSYT and RPP of skew shapes, our analysis of the Hillman-Grassl correspondence gives the following simultaneous generalizations of Gansner's theorem and our theorems 1.4 and 1.5.
As with the (1.5), the RHS of (1.8) can be stated completely in terms of excited diagrams (see Corollary 6.19).
Let us emphasize that the proof Theorem 1.8 requires both the algebraic proof of Theorem 1.4 and the analysis of the Hillman-Grassl correspondence.
1.6.Enumerative applications.In sections 8 and 9, we give enumerative formulas which follow from NHLF.They involve q-analogues of Catalan, Euler and Schröder numbers.We highlight several of these formulas.
Consider the following two q-analogues of E n : where maj(σ) is the major index of permutation σ in S n and κ is the permutation κ = (13254 . ..).See examples 8.6 and 9.4 for the initial values.Now, for the skew shape δ n+2 /δ n , Theorem 1.4 gives the following q-analogue of Corollary 1.9.
Similarly, Theorem 1.5 in this case gives a different q-analogue.
Corollary 1.11.We have: , where and HP(γ) denotes the set of peaks (c, d) in γ with height d > 1.
All three corollaries are derived in sections 8 and 9.
1.7.Comparison with other formulas.In Section 10 we provide a comprehensive overview of the other formulas for f λ/µ that are either already present in the literature or could be deduced.We show that the NHLF is not a restatement of any of them, and in particular demonstrate how it differs in the number of summands and the terms themselves.The classical formulas are the Jacobi-Trudi identity, which has negative terms, and the expansion of f λ/µ via the Littlewood-Richardson rule as a sum over f ν for ν n.Another formula is the Okounkov-Olshanski identity summing particular products over SSYTs of shape µ.While it looks similar to the NHLF, it has more terms and the products are not over hook-lengths.
We outline another approach to formulas for f λ/µ .We observe that the original proof of Naruse of the NHLF in [Naru] comes from a particular specialization of the formal variables in the evaluation of equivariant Schubert structure constants (generalized Littlewood-Richardson coefficients) corresponding to Grassmannian permutations.Ikeda-Naruse give a formula for their evaluation in [IN1] via the excited diagrams on one-hand and an iteration of a Chevalley formula on the other hand, which gives the correspondence with skew standard Young tableaux.Our algebraic proof of Theorem 1.3 follows this approach.Now, there are other expressions for these equivariant Schubert structure constants, which via the above specialization would give enumerative formulas for f λ/µ .First, the Knutson-Tao puzzles [KT] give an enumerative formula as a sum over puzzles of a product of weights corresponding to it.As shown in [MPP] this formula is equivalent to the Okounkov-Olshanski formula and hence different from the sum over excited diagrams.Yet another rule for the evaluation of these specific structure constants is given by Thomas and Yong in [TY], as a sum over certain edge-labeled skew SYTs of products of weights (corresponding to the edge label's paths under jeu-de-taquin).An example in Section 10 illustrates that the terms in the formula are different from the terms in the NHLF.
1.8.Paper outline.The rest of the paper is organized as follows.We begin with notation, basic definitions and background results (Section 2).The definition of excited diagrams is given in Section 3, together with the original formula of Naruse and corollaries of the q-analogue.It also contains the enumerative properties of excited diagrams and the correspondence with flagged tableaux.In Section 4, we give an algebraic proof of the main Theorem 1.4.Section 7 described the Hillman-Grassl correspondence, with various properties and an equivalent formulation using the RSK correspondence in Corollary 5.8.
Section 6 defines pleasant diagrams and proves Theorem 1.5 using the Hillman-Grassl correspondence, and as a corollary gives a purely combinatorial proof of NHLF (Theorem 1.2).Then, in Section 7, we show that the Hillman-Grassl map is a bijection between skew SSYT of shape λ/µ and certain integer arrays whose support is in the complement of an excited diagram.Section 8 considers the special case when λ/µ is a thick strip shape, which give the connection with Euler and Catalan numbers.
In Section 9, we consider the pleasant diagrams of the thick strip shapes, establishing connection with Schröder numbers.We also state conjectures on certain determinantal formulas.Section 10 compares NHLF and other formulas for f λ/µ .We conclude with final remarks and open problems in Section 11.
A skew shape is denoted by λ/µ.For an integer k, 1 Given the skew shape λ/µ, let P λ/µ be the poset of cells (i, j) of [λ/µ] partially ordered by component.This poset is naturally labelled, unless otherwise stated.

Young tableaux.
A reverse plane partition of skew shape λ/µ is an array π = (π ij ) of nonnegative integers of shape λ/µ that is weakly increasing in rows and columns.We denote the set of such plane partitions by RPP(λ/µ).A semistandard Young tableau of shape λ/µ is a RPP of shape λ/µ that is strictly increasing in columns.We denote the set of such tableaux by SSYT(λ/µ).A standard Young tableau (SYT) of shape λ/µ is an array T of shape λ/µ with the numbers 1, . . ., n, where n = |λ/µ|, each i appearing once, strictly increasing in rows and columns.For example, there are five SYT of shape (32/1): The size of a RPP or tableau T is the sum of its entries.A descent of a SYT T is an index i such that i + 1 appears in a row below i.The major index tmaj(T ) is the sum i over all the descents of T .
2.4.Permutations.We write permutations of {1, 2, . . ., n} in one-line notation: w = (w 1 w 2 . . .w n ) where w i is the image of i.A descent of w is an index i such that w i > w i+1 .The major index maj(w) is the sum i of all the descents i of w.

Dyck paths.
A Dyck path γ of length 2n is a lattice paths from (0, 0) to (2n, 0) with steps (1, 1) and (1, −1) that stay on or above the x-axis.We use Dyck(n) to denote the set of Dyck paths of length 2n.For a Dyck path γ, 2.6.Bijections.To avoid ambiguity, we use the word bijection solely as a way to say that map φ : X → Y is one-to-one and onto.We use the word correspondence to refer to an algorithm defining φ.Thus, for example, the Hillman-Grassl correspondence Ψ defines a bijection between certain sets of tableaux and arrays.

Excited diagrams
3.1.Definition and Naruse's formula.Let λ/µ be a skew partition and D be a subset of the Young diagram of λ.A cell u = (i, j) ∈ D is called active if (i + 1, j), (i, j + 1) and (i + 1, j + 1) are all in [λ] \ D. Let u be an active cell of D, define α u (D) to be the set obtained by replacing (i, j) in D by (i + 1, j + 1).We call this replacement an excited move.An excited diagram of λ/µ is a subdiagram of λ obtained from the Young diagram of µ after a sequence of excited moves on active cells.Let E(λ/µ) be the set of excited diagrams of λ/µ.
Next, by [S3,Cor. 7.21.3] the principal specialization of the Schur function s (k,1 d−1 ) equals where n k q is a q-binomial coefficient.Second if the excited diagram D ∈ E(λ/µ) corresponds to path γ then one can show that a(D) = n 2 + area(γ) where area(γ) is the number of cells in the d × k rectangle South East of the path γ.Putting this all together then Theorem 1.4 for shape λ/µ gives In [MPP], we show that (3.1) and (3.2) are special cases of Racah and q-Racah formulas in [BGR].
Next, we show that Theorem 1.4 is a q-analogue of (1.2).This argument is standard; we outline it for reader's convenience.
Theorem 1.5 is a different q-analogue of NHLF, as explained in Section 6.
3.2.Flagged tableaux.Excited diagrams of λ/µ are also equivalent to certain flagged tableaux of shape µ (see Proposition 3.7 and [Kre1,§6]) and thus the number of excited diagrams is given by a determinant (see Corollary 3.8), a polynomial in the parts of λ and µ.
In this section we relate excited diagrams with flagged tableaux.The relation is based on a map by Kreiman [Kre1,§6] (see also [KMY,§5]).
We start by stating an important property of excited diagrams that follows immediately from their construction.Given a set D ⊆ [λ] we say that (i, j), (i + m, j + m) ∈ D ∩ d k for m > 0 are consecutive if there is no other element in D on diagonal d k between them.Fix a sequence f = (f 1 , f 2 , . . ., f (µ) ) of nonnegative integers.Define F(µ, f ) to be the set of T ∈ SSYT(µ), such that all entries T ij ≤ f i .Such tableaux are called flagged SSYT and they were first studied by Lascoux and Schützenberger [LS] and Wachs [Wac].By the Lindström-Gessel-Viennot lemma on non-intersecting paths (see e.g.[S3,Thm. 7.16.1]), the size of F(µ, f ) is given by a determinant: Proposition 3.6 (Gessel-Viennot [GV], Wachs [Wac]).In the notation above, we have: , where h k (x 1 , x 2 , . ..) denotes the complete symmetric function.
Given a skew shape λ/µ, each row i of µ is between the rows k i−1 < i ≤ k i of two corners of µ.When a corner of µ is in row k, let f k be the last row of diagonal d µ k −k in λ.Lastly, let f (λ/µ) be the vector1 (f 1 , f 2 , . . ., f (µ) ), f i = f ki where k i is the row of the corner of µ at or immediately after row i (see Figure 2).Let F(λ/µ) := F(µ, f (λ/µ) ).
Let T µ be the tableaux of shape µ with entries i in row i.Note that T µ ∈ F(λ/µ).We define an analogue of an excited move for flagged tableaux.A cell (x, y) of T in F(λ/µ) is active if increasing T x,y by 1 results in a flag SSYT tableau T in F(λ/µ).We call this map T → T a flagged move and denote by α x,y (T ) = T .
Next we show that excited diagrams in E(λ/µ) are in bijection with flagged tableaux in F(λ/µ).
Given D ∈ E(λ/µ), we define ϕ(D) := T as follows: Each cell (x, y) of [µ] corresponds to a cell (i x , j y ) of D. We let T be the tableau of shape µ with T x,y = i x .An example is given in Figure 2.
Figure 2. Given a skew shape λ/µ, for each corner k of µ we record the last row f k of λ from diagonal d µ k −k .These row numbers give the bound for the flagged tableaux of shape µ in F(µ, f (λ/µ) ).
By Proposition 3.6, we immediately have the following corollary.
Since the correspondences ϕ from Propositions 3.7 and 3.9 are the same then both sets of tableaux are equal.
Remark 3.11.To clarify the unusual situation in this section, here we have three equinumerous sets K(λ/µ), F(λ/µ) and E(λ/µ), all of which were previously defined in the literature.The first two are in fact the same sets, but defined somewhat differently; essentially, the set of inequalities in the definition of K(λ/µ) is redundant.Since our goal is to prove Corollary 3.8, we find it easier and more instructive to use Kreiman's map ϕ with a new analysis (see below), to prove directly that |E(λ/µ)| = |F(λ/µ)|.An alternative approach would be to prove the equality of sets F(λ/µ) = K(λ/µ) first (Corollary 3.10), which reduces the problem to Kreiman's result (Proposition 3.9).
Proof of Proposition 3.7.We need to prove that ϕ is a well defined map from E(λ/µ) to F(µ, f (λ/µ) ).First, let us show that T = ϕ(D) is a SSYT by induction on the number of excited moves of D. First, note that ϕ([µ]) = T µ which is SSYT.Next, assume that for D ∈ E(λ/µ), T = ϕ(D) is a SSYT and D = α (ix,jy) (D) for some active cell (i x , j y ) of D corresponding to (x, y) in [µ].Then T = ϕ(D ) is obtained from T by adding 1 to entry T x,y = i x and leaving the rest of entries unchanged.When (x + 1, y) ∈ [µ], since (i x + 1, j y ) is not in D then the cell of the diagram corresponding to (x + 1, y) is in a row > i x + 1, therefore T x,y = i x + 1 < T x+1,y = T x+1,y .Similarly, if (x, y + 1) ∈ [µ], since (i x , j x +1) is not in D then the cell of the diagram corresponding to (x, y +1) is in a row > i x , therefore T x,y = i x + 1 ≤ T x,y+1 = T x,y+1 .Thus, T ∈ SSYT(λ/µ).
Next, let us show that T is a flagged tableau in F(µ, f (λ/µ) ).Given an excited diagram D, if cell (i x , j y ) of D is the cell corresponding to (x, y) in [µ] then the row i x is at most f kx : the last row of diagonal d µ kx −kx where k x is the row of the corner of µ on or immediately after row x.Thus T x,y ≤ f kx , which proves the claim.
Finally, we prove that ϕ is a bijection by building its inverse.Given T ∈ F(µ, f (λ/µ) ), let D = ϑ(T ) be the set D = {(T x,y , y + T x,y ) | (x, y) ∈ [µ]}.Let us show ϑ is a well defined map from F(µ, f (λ/µ) ) to E(λ/µ).By definition of the flags f (λ/µ) , observe that D is a subset of [λ].We prove that D is in E(λ/µ) by induction on the number of flagged moves α x,y (•).First, observe that ϑ(T µ ) = [µ] which is in E(λ/µ).Assume that for T ∈ F(λ/µ), D = ϑ(T ) is in E(λ/µ) and T = α x,y (T ) for some active cell (x, y) of T .Note that replacing T x,y by T x,y + 1 results in a flagged tableaux T in F(λ/µ) is equivalent to (i x , i y ) being an active cell of D. Since ϑ(T ) = α ix,iy (D) and the latter is an excited diagram, the result follows.By construction, we conclude that ϑ = ϕ −1 , as desired.Note that v(1), . . ., v(n . ., λ 1 + d} arranged in increasing order.The numbers written up to the vertical segment on row i are 1, . . ., λ i + d − i, of which d − i are on the first vertical segments, and the other λ i are on the first horizontal segments.This gives We use results from Ikeda and Naruse [IN1].Let [X w ] be the Schubert class corresponding to a permutation w and let [X w ]| v be the multivariate polynomial with variables y 1 , . . ., y n corresponding to the image of the class under a certain homomorphism ι v . Theorem 4.1 (Theorem 1 in [IN1], Prop.2.2 (ii) in [Kre1]).Let w v be Grassmannian permutations whose unique descent is at position [KK, Bil, Tym].Billey's formula [AJS,Appendix D.3], [Bil,Eq. (4.5)] expresses the latter as certain sums over reduced subwords of w from a fixed reduced word of v. Since in our context w and v are Grassmannian, the reduced subwords are related only by commutations and no braid relations (cf.[Ste]).This property allows the authors in [IN1] to find a bijection between the reduced subwords and excited diagrams.The author in [Kre1] uses the different method of Gröbner degenerations to prove the result.
Corollary 4.4.Let w v be Grassmannian permutations whose unique descent is at position Proof.Combining Theorem 4.1 and Theorem 4.3 we get Note that (w) = |µ| and (v) = |λ|, so we can remove the (−1) (w) on the left of (4.2) by negating all linear terms on the right and get the desired result.4.2.Proof of Theorem 1.4.First we use Corollary 4.4 to get an identity of rational functions in y = (y 1 , y 2 , . . ., y n ) (Lemma 4.5).Then we evaluate this identity at y p = q p−1 and use some identities of symmetric functions to prove the theorem.Let .
Denote the LHS of (4.5) by S λ,µ (y).By the determinantal formula for factorial Schur functions and by (4.4) we have .
Using ( * * ) in the denominator of the matrix entry, we obtain: Therefore, the matrix entry on the RHS of (4.6) simplifies to H i,r (y).
where h k (x) denotes the k-th complete symmetric function.
Proof.We have: where the last identity follows by the principal specialization of the complete symmetric function.
It remains to match the powers of q in (4.11) and (1.4).
Proposition 4.7.For an excited diagram D ∈ E(λ/µ) we have: Finally, notice that the cells of any excited diagram D have the same multiset of content values, since every excited move is along a diagonal and the content of the moved cell j − i remains constant.Thus the power of q for each term becomes as desired.
Using Proposition 4.7 on the RHS of (4.11) yields (1.4) finishing the proof of Theorem 1.4.
5. The Hillman-Grassl and the RSK correspondences 5.1.The Hillman-Grassl correspondence.Recall the Hillman-Grassl correspondence which defines a map between RPP π of shape λ and arrays A of nonnegative integers of shape λ such that We review this construction and some of its properties (see [S3, §7.22] and [Sag2, §4.2]).We denote by Φ the Hillman-Grassl map Φ : π → A.
Definition 5.1 (Hillman-Grassl map Φ).Given a reverse plane partition π of shape λ, let A be an array of zeroes of shape λ.Next we find a path p of North and East steps in π as follows: (i) Start p with the most South-Western nonzero entry in π.Let c s be the column of such an entry.(ii) If p has reached (i, j) and π i,j = π i−1,j > 0 then p moves North to (i − 1, j), otherwise if 0 < π i,j < π i−1,j then p moves East to (i + 1, j). (iii) The path p terminates when the previous move is not possible in a cell at row r f .Let π be obtained from π by subtracting 1 from every entry in p.Note that π is still a RPP.In the array A we add 1 in position A cs,r f and obtain array A .We iterate these three steps until we reach a plane partition of zeroes.We map π to the final array A.
Definition 5.3 (Inverse Hillman-Grassl map Φ −1 ).Given an array A of nonnegative integers of shape λ, let π be the RPP of shape λ of all zeroes.Next, we order the nonzero entries of A, counting multiplicities, with the order (i, j) < (i , j ) if j > j or j = j and i < i (i.e.(i, j) is right of (i , j ) or higher in the same column).Next, for each entry (r s , c j ) of A in this order (i 1 , j 1 ), . . ., (i m , j m ) we build a reverse path q of South and West steps in π starting at row r s and ending in column c f as follows: (i) Start q with the most Eastern entry of π in row r s .
(ii) If q has reached (i, j) and π i,j = π i+1,j then q moves South to (i − 1, j), otherwise q moves West to (i + 1, j). (iii) Path q ends when it reaches the Southern entry of π in column c f .
Step (iii) is actually attained (see e.g.[Sag2,Lemma 4.2.4].Let π be obtained from π by adding 1 from every entry in q.Note that π is still a RPP.In the array A we subtract 1 in position A c f ,rs and obtain array A .We iterate this process following the order of the nonzero entries of A until we reach an array of zeroes.We map A to the final RPP π.Note that ω(A) = |π|.
By abuse of notation, if π is a skew RPP of shape λ/µ, we define Φ(π) to be Φ(π) where π is the RPP of shape λ with zeroes in µ and agreeing with π in λ/µ: Recall that unlike for straight shapes, the enumeration of SSYT and RPP of skew shape are not equivalent.Therefore, the image Φ(SSYT(λ/µ)) is a strict subset of Φ(RPP(λ/µ)).In Section 7 we characterize the SSYT case in terms of excited diagrams, and in Section 6 we characterize the RPP case in terms of new diagrams called pleasant diagrams.Both characterizations require a few properties of Φ that we review next.
5.2.The Hillman-Grassl correspondence and Greene's theorem.In this section we review key properties of the Hillman-Grassl correspondence related to the RSK correspondence [S3, §7.11].We denote Ψ : M → (P, Q), where M is a matrix with nonnegative integer entries and I(Ψ(M )) := P , R(Ψ(M )) := Q are SSYT of the same shape called the insertion and recording tableau, respectively.Given a reverse plane partition π and an integer k with 1 of π is nonincreasing and so we denote it by a partition ν (k) .The k-trace of π denoted by tr k (π) is the sum of the parts of ν (k) .Note that the 0-trace of π is the standard trace tr(π) = i π i,i .
Given the Young diagram of λ and an integer k with 1 − (λ) ≤ k ≤ λ 1 − 1, let λ k be the largest i × (i + k) rectangle that fits inside the Young diagram starting at (1, 1).For k = 0, the rectangle λ 0 = λ is the (usual) Durfee square of λ.Given an array A of shape λ, let A k be the subarray of A consisting of the cells inside λ k and |A k | be the sum of its entries.Also, given a rectangular array B, let B and B ↔ denote the arrays B flipped vertically and horizontally, respectively.Here vertical flip means that the bottom row become the top row, and horizontal means that the rightmost column becomes the leftmost column.
In the construction Φ −1 , entry 1 in position (i, j) adds 1 to the k-trace if and only if (i, j) ∈ λ k .This observation implies the following result.
As a corollary, when k = 0, Proposition 5.5 gives Gansner's formula (1.7) for the generating series for RPP(λ) by size and trace.Indeed, the generating function for the arrays is a product over cells (i, j) ∈ [λ] of terms which contain t in the numerator if only if (i, j) ∈ λ .We refer to [G1] for the details.
Let us note that not only is the k-trace determined by Proposition 5.5 but also the parts of ν (k) .This next result states that the partition ν (k) and its conjugate are determined by nondecreasing and nonincreasing chains in the rectangle A k .
Given an m × n array M = (m ij ) of nonnegative integers, an ascending chain of length s of M is a sequence c := ((i 1 , j 1 ), (i 2 , j 2 ), . . ., (i s , j s )) where Let ac 1 (M ) and dc 1 (M ) be the length of the longest ascending and descending chains in M respectively.In general for t ≥ 1, let ac t (M ) be the maximum combined length of t ascending chains where the combined number of times (i, j) appears is m ij .We define dc t (M ) analogously for descending chains.

Hillman-Grassl map on skew RPP
In this section we show that the Hillman-Grassl map is a bijection between RPP of skew shape and arrays of nonnegative integers with support on certain diagrams related to excited diagrams.
6.1.Pleasant diagrams.We identify any diagram S (set of boxes in [λ]) with its corresponding 0-1 indicator array, i.e. array of shape λ and support S.
the subarray S k := S ∩ λ k has no descending chain bigger than the length s k of the diagonal d k of λ/µ, i.e. for every k we have dc 1 (S k ) ≤ s k .We denote the set of pleasant diagrams of λ/µ by P(λ/µ).
Example 6.2.The skew shape (2 2 /1) has 12 pleasant diagrams of which two are complements of excited diagrams (the first in each row): . These are diagrams S of [2 2 ] where S ∩ λ −1 , S ∩ λ 0 and S ∩ λ 1 have no descending chain bigger than Theorem 6.3.A RPP π of shape λ has support in a skew shape λ/µ if and only if the support of Φ(π) is a pleasant diagram in P(λ/µ).In particular (6.1) Proof.By Theorem 5.6, a RPP π of shape λ has support in the skew shape λ/µ if and only if k) .In other words, π has support in the skew shape λ/µ if and only if the support S ⊆ [λ] of A is in P(λ/µ).Thus, the Hillman-Grassl map is a bijection between RPP(λ/µ) and arrays of nonnegative integers of shape λ with support in a pleasant diagram S ∈ P(λ/µ).This proves the first claim.Equation (6.1) follows since |π| = ω(Φ(π)).Remark 6.4.Theorem 1.5 gives an alternative description for pleasant diagrams P(λ/µ) as the supports of 0-1 arrays A of shape λ such that Φ −1 (A) is in RPP(λ/µ).
We also give a generalization of the trace generating function (1.7) for these reverse plane partitions.
Proof of Theorem 1.7.Given a pleasant diagram S ∈ P(λ/µ), let B S be the collection of arrays of shape λ with support in S. Given a RPP π, let A = Φ(π).By Theorem 6.3 π has shape λ/µ if and only if A has support in a pleasant diagram S in P(λ/µ).Thus (6.2) where for each S ∈ P(λ/µ) we have (6.3) Next, by Proposition 5.5 for k = 0, the trace tr(π) equals |A 0 |, the sum of the entries of A in the Durfee square λ of λ.Therefore, we refine (6.3) to keep track of the trace of the RPP and obtain (6.4) Combining (6.2) and (6.4) gives the desired result.6.2.Combinatorial proof of NHLF (1.2): relation between pleasant and excited diagrams.Theorem 1.4 relates SSYT of skew shape with excited diagrams and Theorem 6.3 relates RPP of skew shape with pleasant diagrams.Since SSYT are RPP then we expect a relation between pleasant and excited diagrams of a fixed skew shape λ/µ.The first main result of this subsection characterizes the pleasant diagrams of maximal size in terms of excited diagrams.The second main result of the next subsection characterizes all pleasant diagrams.
The key towards these results is a more graphical characterization of pleasant diagrams as described in the proof of Lemma 6.6.It makes the relationship with excited diagrams more apparent and also allows for a more intuitive description for both kinds of diagrams.By combining this theorem with Theorem 6.3 we derive again the NHLF.In contrast with the derivation of this formula in Proposition 3.4, this derivation is entirely combinatorial.
Proof.Given a pleasant diagram S, we use Viennot's shadow lines construction [Vie] to obtain a family of nonintersecting paths on [λ].That is, we imagine there is a light source at the (1, 1) corner of [λ] and the elements of S cast shadows along the axes with vertical and horizontal segments.The boundary of the resulting shadow forms the first shadow line L 1 .If lines L 1 , L 2 , . . ., L i−1 have already been defined we define L i inductively as follows: remove the elements of S contained in any of the i − 1 lines and set L i to be the shadow line of the remaining elements of S. We iterate this until no element of S remains in the shadows.Let L 1 , L 2 , . . ., L m be the shadow lines produced.Note that these lines form m nonintersecting paths in [λ] that go from bottom south-west cells of columns to rightmost north-east cells of rows of the diagram.By construction the peaks (i.e.top corners) of the shadow lines L i are elements of S while other cells of L i might be in [λ] \ S.
Next we augment S to obtain S * by adding all the cells of lines L 1 , . . ., L m that are not in S. Note that S * is also a pleasant diagram in P(λ/µ) since the added cells of the lines L 1 , . . ., L m do not yield longer decreasing chains than those in S.Moreover, no two cells from a decreasing chain can be part of the same shadow line, and there is at least one decreasing subsequence with cells in all lines, as can be constructed by induction.In particular, the number of shadow lines intersecting each diagonal d k (i.e.intersecting the rectangle λ k ) is at most s k .Denote this number by s k .Next, we claim that S * is the complement of an excited diagram D * ∈ E(λ/ν) for some partition ν.
To see this we do moves on the noncrossing paths (shadow lines) that are analogous to reverse excited moves, as follows.If the lines contain (i, j), (i + 1, j), (i, j + 1) but not (i + 1, j + 1), then notice that the first three boxes lie on one path L t .In this path we replace (i, j) with (i + 1, j + 1) to obtain path L t .We do the same replacement in S * : Since a diagonal of length has exactly shadow lines crossing it, we have for each rectangle D k there are exactly s k lines L i crossing d k and hence also crossing D k .An excited move corresponds to a ladder move on some line (see the proof of Lemma 6.6), which makes an inner corner of a line into an outer corner.These moves cannot affect the endpoints of a line, so if a line does not cross a rectangle D k initially then it will never cross it after any number of excited moves.Moreover, any diagonal d k will be crossed by the same set of lines formed originally in P 0 .Hence the complement of any excited diagram is a collection of shadow lines, which were obtained from the original ones by ladder moves.Then the number of shadow lines crossing D k is always s k .Finally, since no decreasing sequence can have more than one box on a given shadow line (i.e., a SW to NE lattice path), we have the longest decreasing subsequence in D k will have length at most s k -the number of shadow lines there.Therefore, the excited diagram satisfies Definition 6.1.By Lemma 6.7, the complements of excited diagrams in E(λ/µ) give pleasant diagrams of size |λ/µ|.Next, we show that there are no pleasant diagrams of larger size.Lemma 6.8.For all S ∈ P(λ/µ), we have |S| ≤ |λ/µ|.
Proof.For each diagonal d k of λ/µ, any elements of S ∩ d k form a descending chain in S k .Thus, by definition of pleasant diagrams Proof of Theorem 6.5.The result follows by combining Lemmas 6.7 and 6.9.We prove this result via two lemmas.Lemma 6.11.
Proof.Theorem 6.5 characterizes maximal pleasant diagrams in P(λ/µ) as complements of excited diagrams in E(λ/µ).Since subsets of pleasant diagrams are also pleasant diagrams, then all subsets S of [λ] \ D for D ∈ E(λ/µ) are pleasant diagrams.
Proof of Theorem 6.10.The theorem follows from Lemma 6.11 and Lemma 6.6.
Next we give two formulas for the number of pleasant diagrams of λ/µ as sums of excited diagrams.Both formulas are corollaries of the proof of Lemma 6.6.Given a pleasant diagram S, let shpk(D) be the number of peaks of the shadow lines L 1 , . . ., L m obtained from the pleasant diagram [λ] \ D. Proposition 6.12.
Proof of Proposition 6.12.As in the proof of Lemma 6.6, from the shadow lines L 1 , L 2 , . . ., L m of a pleasant diagram S ∈ P(λ/µ) we obtain an excited diagram D * ∈ E(λ/ν) for µ ⊆ ν such that S ⊆ [λ] \ D * .The peaks of these lines are elements in S, and these peaks uniquely determine the lines.The other cells in the lines, |λ/ν| − shpk(D * ) many, may or may not be in S.
Therefore, we obtain a surjection . This implies the result (see Figure 6).
That is, the excited peaks of α u (D) are obtained from those of D by adding (i, j) and removing (i, j +1) and (i + 1, j) if any of the two are in Λ(D): . We prove Theorem 6.14 via the following Lemma.Given a set S, let 2 S denote the subsets of S. Lemma 6.15.We have P(λ/µ) = D∈E(λ/µ) Λ(D) × 2 [λ]\(D∪Λ(D)) .
Proof.As in the proof of Lemma 6.6, from the shadow lines L 1 , L 2 , . . ., L m of a pleasant diagram S ∈ P(λ/µ) we obtain an excited diagram D * ∈ E(λ/ν) for µ ⊆ ν such that S ⊆ [λ] \ D * .If we restrict D * to the cells coming from [µ] we obtain an excited diagram D ∈ E(λ/µ).Setting 2 (S) = D defines a new surjection 2 : P(λ/µ) → E(λ/µ) (see Figure 6).It remains to prove that First, the excited peaks are peaks of the shadow lines L 1 , L 2 , . . ., L k of [λ] \ D obtained by a ladder move: Thus the peaks of the shadow lines {L i } are either excited peaks or original peaks of the shadow lines of [λ/µ].Second, note that the excited peaks Λ(D) determine uniquely the excited diagram D. Thus the non-excited peaks of the shadow lines and the other cells of the lines {L i }, those in [λ]\(D ∪Λ(D)), may or may not be in S.This proves the claim for −1 2 (D).
6.4.Excited diagrams and skew RPP.In Section 6.1 we expressed the generating function of skew RPP using pleasant diagrams.In this section we use Lemma 6.15 to give an expression for this generating series in terms of excited diagrams.
Proof of Corollary 6.17.By Theorem 6.3, we have: Using Lemma 6.15 and the surjection ϑ 2 in its proof, we can rewrite the RHS above as a sum over excited diagrams.We have: This result also implies the NHLF (1.2).

Hillman-Grassl map on skew SSYT
In this section we show that the Hillman-Grassl map is a bijection between SSYT of skew shape and certain arrays of nonnegative integers with support in the complement of excited diagrams and some forced nonzero entries.First, we describe these arrays and state the main result.
Definition 7.1 (excited arrays).For an excited diagram D in E(λ/µ) obtained from [µ] by a sequence of excited moves the operations β u are well defined.So each excited diagram D is associated to a 0-1 array A D (see Figure 8).
Next we show that the procedure for obtaining the arrays A D is well defined; meaning that at each stage, the conditions to apply β u (•) are met.
) is also an active cell of D, then the excited move β u (•) did not alter the values at (i + 1, j + 1) and (i + 1, j ).In this case (A D ) i +1,j +1 = (A D ) i+1,j+1 = 1 and (A D ) i +1,j = (A D ) i +1,j = 0.If v is not an active square of D then u is one of {(i , j + 1), (i + 1, j ), (i − 1, j − 1)} (note that u = (i + 1, j + 1) since the corresponding flagged tableau would not be semistandard).In each of these three cases we see that (A D ) i +1,j +1 = 1 and (A D ) i +1,j = 0: Remark 7.4.Each broken diagonal d t (D) is a sequence of diagonal segments from d t (µ) broken by horizontal segments coming from row µ t .We call these segments excited segments.
Definition 7.6.For D ∈ E(λ/µ), let A * D be the set of arrays A of nonnegative integers of shape λ with support contained in [λ] \ D, and nonzero entries A i,j > 0 if (A D ) i,j = 1, where A D is 0-1 excited array corresponding to D.
We are now ready to state the main result of this section.
Theorem 7.7.The (restricted) Hillman-Grassl map Φ is a bijection: We postpone the proof until later in this section.Let us first present the applications of this result.Note first that since Φ(•) is weight preserving, Theorem 7.7 implies an alternative description of the statistic a(D) = u∈D (λ j − i) from (1.4) in terms of sums of hook-lengths of the support of A D (i.e. the weight ω(A D )).
Since by Theorem 1.5 we understand the image of the Hillman-Grassl map on SSYT of skew shape then we are able to give a generalization of the trace generating function (1.7) for these SSYT.
Proof of Theorem 1.8.By Theorem 7.7 a tableau T has shape λ/µ if and only if where for each D ∈ E(λ/µ) we have: Next, by Proposition 5.5 for k = 0, the trace tr(π) equals |A 0 |, the sum of the entries of A in the Durfee square λ of λ.Therefore, we refine (7.3) to keep track of the trace of the SSYT and obtain (7.4) where c(D) = |supp(A D ) ∩ λ | and ω(A D ) = a(D).Combining (7.2) and (7.4) gives the result.
Proof of Theorem 7.7: First we use Theorem 6.3 to show that Φ −1 ( U ∈E(λ/µ) A * D ) consists of RPP of skew shape λ/µ (Lemma 7.10).Then we show that these RPP are also column-strict (Lemma 7.11).These two results and the fact that Φ −1 is injective imply that In addition, since Φ is weight preserving, we have: where By Theorem 1.4 and the equality a(D) = ω(A D ) (Proposition 7.16), it follows that the difference in (7.5) is zero.Therefore, the restricted map Φ is a bijection between tableaux in SSYT(λ/µ) and arrays in U ∈E(λ/µ) A * D , as desired.Let π be the reverse plane partition Φ −1 (A) for A ∈ A * D and D ∈ E(λ/µ).By Lemma 7.10, we know that π has support in the skew shape λ/µ.We show that π has strictly increasing columns by comparing any two adjacent entries from the same column of π .Consider the two adjacent diagonals of π to which the corresponding entries belong and let ν 1 and ν 2 be the partitions obtained by reading these diagonals bottom to top.There are two cases depending on whether the diagonals end in the same column or in the same row of λ/µ; Case 1: If the diagonals end in the same column, then it suffices to show that ν 2 i < ν 1 i for all i.Case 2: If the diagonals end in the same row, then it suffices to show that ν 2 i+1 < ν 1 i for all i.

w v
Before we treat these cases we prove the following Lemma needed for both.
Lemma 7.12.Let M be a rectangular array coming from A ∈ A * D with NW corner (1, 1).Then the first column of P = I(Ψ(M )) is (1, . . ., h), where h is the height of P .
Proof.We will use the symmetry of the RSK correspondence.Recall that Ψ(N ) = (P, Q) for some rectangular array N then Ψ(N T ) = (Q, P ) so that P is the recording tableaux by doing RSK on N row by row, bottom to top.Thus the first column of P gives the row numbers of N where the height of the insertion tableaux increased by one.
Let R be the rectangular shape of M .By Greene's theorem, h is equal to the length of the longest decreasing subsequence in M .By Lemma 6.7, h is at most the length of longest diagonal of R/µ.
Note that M contains a broken diagonal of length at least h − 1 since either the longest diagonal of length h in R/µ ends in a vertical step of µ, in which case M has a broken diagonal of the same length, or the longest diagonal ends in a horizontal step of µ in which case M has a broken diagonal of length h − 1.
Let d be such a broken diagonal.Since a broken diagonal is a decreasing subsequence that spans consecutive rows, then d spans the lower h − 1 rows of M .This guarantees that the first column of P is 1, 2, . . ., h − 1, c, where c ≥ h is the row where we first get a decreasing subsequence of length h.
Assume there is a longest decreasing subsequence d of length h whose first cell x = (i 1 , j 1 ) is in a row c = i 1 > h (counting rows bottom to top), and take both i 1 and j 1 to be minimal.
Either x is inside or outside of [µ].If x is outside then there is a diagonal that ends in row i 1 − 1 to the left of x, which results in a broken diagonal of length i 1 − 1 ≥ h in the excited diagram.Hence, . Two cases to consider in the proof of Lemma 7.12 depending on whether cell x = (i 1 , j 1 ) is outside or inside of [µ].
there is a decreasing subsequence of length h starting at a lower row than the row i 1 of x, leading to a contradiction.See Figure 10 : Left.
When x is inside of [µ] then there is an excited cell below x in the same diagonal.There must be a broken diagonal d that reaches at least row i 1 − 1 below or to the left of x.At row i 1 , the sequence d is above d and the last entry of d is below d , as otherwise d would be shorter than d .Thus the sequence d and the broken diagonal d cross.Consider the first crossing tracing top down.Let a be the last cell of d before this crossing and let b be the cell of d on or after the crossing.Note that below a in the same column there is either a nonzero from d or a zero from the excited horizontal segment of d .In either case, a is higher than the lowest cell of d to the left of b.Define d to be the sequence consisting of the segment of d from row i 1 − 1 up until the crossing followed by the segment of d from cell b onwards (see Figure 10: Right).Note that d is a decreasing sequence of R that starts at row i 1 − 1 and column ≤ j 1 and has length h since d includes a nonzero element from the row below the row a.This contradicts the minimality of x.
In summary, we conclude that c = h, and the first column of P is (1, . . ., h), as desired.This finishes the proof.
Column strictness in Case 1.By Corollary 5.8 we have ν 1 = shape(P 1 ) and ν 2 = shape(P 2 ) where P 1 = I(Ψ(M 1 )), P 2 = I(Ψ(M 2 )), and the rectangular array M 1 = A t ↔ is obtained from the rectangular array M 2 = A t+1 ↔ by adding a row w at the end.Thus ν 1 is the shape of the insertion tableau obtained by row inserting w (from left to right) in the insertion tableau I(Ψ(M 2 )) of shape ν 2 .Proposition 7.13.In Case 1 we have ν 2 i < ν 1 i for 1 ≤ i ≤ min{ (ν 1 ), (ν 2 )}.Proof.Let P = I(Ψ(M 1 )) and Q = R(Ψ(M 1 )) (in this case, M 1 is being read top to bottom, left to right, i.e. row by row starting from the right, from the original array A t before the flip).Let m be the height of M 1 .The strict inequality is equivalent to the fact that the insertion of the last row in P results in an extension of every row, i.e. every row of the recording tableau Q has at least one entry equal to m.We will prove the last statement.Note that by the symmetry of the RSK correspondence, we have Q is the insertion tableaux for A t when read column by column from right to left.
Claim: Let h be the height of Q, i.e. the longest decreasing subsequence of M 1 .Then row i of Q contains at least one entry from each of m, . . ., m − h + i.
Note that h is equal to the length of the longest broken diagonal or one more than that.We prove the claim by induction on the number of columns in M 1 , i.e. in A t .Let M 1 = [u 1 , u 2 , . . ., u r ], where u i is its i-th column.In terms of the excited array A t , we have that u r is the first column of A t and u 1 -the last.Suppose that the claim is true for A t restricted to its first r − 1 columns, which is still an excited array by definition, and let u = u 1 be its last column.Let Q be the insertion tableaux of [u 2 , . . ., u r ] read column by column, i.e.Q = u 2 ← u 3 ← • • • , where ← indicates the insertion of the corresponding sequence.Then let Q be the insertion tableaux corresponding to A t , so Q = u ← reading(Q) by Knuth equivalence.The reading word is obtained from Q by reading it row by row from the bottom to the top, each row read left to right.
First, suppose that u does not increase the length of the longest decreasing subsequence, so Q has also height h.Let a ∈ [m − j + i, m] be a number present in row i of Q.When it is inserted in Q it will first be added to row 1, where there could be other entries equal to a already present.The first such entry a will be bumped by something ≤ a − 1 coming from inserting row i − 1 of Q into Q .This had to happen in Q since a reached row i.From then on the same numbers will bump each other as in the original insertion which created Q.Thus an entry equal to a will reach row i after the i − 1 bumps.Since the height of Q is unchanged, the claim holds as it pertains only to the original entries a from Q which again occupy the corresponding rows.
Next, suppose that u increases the length of the longest decreasing subsequence to h + 1.Then the longest broken diagonal in A t has length at least h.Also, column u must have an element equal to m, i.e. a nonzero entry in A t 's lower right corner.Moreover, we claim that the longest decreasing subsequence has to occupy the consecutive rows of A t from m−h to m, and thus the longest decreasing subsequences in u, reading(Q) are m, m − 1, . . ., m − h.This is shown within the proof of Lemma 7.12.From there on, in u ← reading(Q) we have element m from u bumped by something ≤ m − 1 from the last row of Q. Afterwards, the bumps happen similarly to the previous case and the numbers from Q reach their corresponding rows, so the m from u reaches eventually one row below, i.e. row h + 1.The entry m − h from the longest decreasing sequence is inserted from the first row of Q and is, therefore, in row 1 of Q , so by iteration Q has the desired structure.This ends the proof of the claim and thus the Proposition.
Column strictness in Case 2. By Corollary 5.8 we have ν 1 = shape(P 1 ) where P 1 = I(Ψ(M 1 )) and ν 2 = shape(P 2 ) where P 2 = I(Ψ(M 2 )) and the rectangular array M 2 = A t+1 is obtained from the rectangular array M 1 = A t by adding a column v at the end (we read column by column SW to NE).Thus ν 2 is the shape of the insertion tableau obtained by row inserting v (from top to bottom) in the insertion tableau I(Ψ(M 1 )) of shape ν 1 .Proposition 7.14.For Case 2 we have ν 2 i+1 < ν 1 i for 1 ≤ i ≤ min{ (ν 1 ), (ν 2 ) − 1}.We prove a stronger statement that requires some notation.Let P be the insertion tableau of shape ν where M = B for some rectangular array B of A ∈ A * D with NW corner (1, 1).for a positive integer k, let P i (k) be the number of entries in row i of P which are ≤ k.
Lemma 7.15.With P and P i (k) as defined above, for k > 1 we have (i Proof of Proposition 7.14.We first show that Lemma 7.15 implies that the insertion path of the RSK map of M moves strictly to the left.To see this, let P be the resulting tableaux obtained at some stage of the insertion when j is inserted in row 1 and bumps j 1 > j to row 2. Then j is inserted at position P 1 (j 1 − 1) in row 1 and j 1 is inserted at position P 2 (j 1 ) > 0 in row 2. By Condition (i), P 2 (j 1 ) < P 1 (j 1 − 1).Iterating this argument as elements get bumped in lower rows implies the claim.
Next, note that a bumped element at position ν 1 2 + 1 from row 1 of P 1 cannot be added to row 2 as otherwise the insertion path would move strictly down, violating Condition (i).Thus the only elements from row 1 of P 1 that can be added to row 2 in P 2 are those in positions > ν 1 2 + 1.And so there are no more than ν 1 1 − ν 1 2 − 1 such elements implying that ν 2 2 ≤ ν 1 1 − 1. Iterating this argument in the other rows implies the result.
Proof of Lemma 7.15.Note that Condition (ii) for k = i + 1 follows by Lemma 7.12.Note that the statement of the lemma holds for any step of the insertion, since it applies for P as a recording tableaux.Since P i (k) are increasing for k with i fixed then Condition (ii) holds.We claim that after each single insertion of Ψ, Condition (i) still holds.We prove this when inserting an element j in row r.Iterating this argument as elements get bumped in lower rows implies the claim.
Assume P verifies Condition (i) and we insert j in row r of P to obtain a tableaux P of shape ν .By Lemma 7.12 we have j ≥ r.If j is added to the end of the row then Condition (i) still holds for P since P r (j) > P r (j).If j bumps j 1 in row r then j 1 > j and (7.6) P r (j) = P r (j 1 − 1) = P r (j 1 − 1) + 1 , P r (j 1 ) = P r (j 1 ) and all other P r (i) = P r (i) remain the same.Next, we insert j 1 in row r + 1 of P .Regardless of whether j 1 is added to the end of the row or bumps another element to row r + 2, we have P r+1 (j) = P r+1 (j) , P r+1 (j 1 ) = P r+1 (j 1 ) + 1 , and all other P r+1 (i) = P r+1 (i) remain the same.Since P r (b) ≥ P r (b) for all b, we need to verify Condition (i) only when P r+1 increased with respect to P r+1 .By Lemma 7.12, we have either row r + 1 of P is nonempty and thus P r+1 (j 1 ) > 0, or else we must have j 1 = r + 1.In the first case Condition (i) applies to P and we have P r+1 (j 1 ) < P r (j 1 − 1).By (7.6), we have P r+1 (j 1 ) = P r+1 (j 1 ) + 1 < P r (j 1 − 1) + 1 = P r (j 1 − 1).Finally, suppose j 1 = r + 1.Since j ≥ r, we then have j = r, and r must have been present in row r in P by Lemma 7.12.Thus P r (r) ≥ 1 and Therefore, Condition (i) is verified for rows r and r + 1 of P , as desired.

Equality between a(D) and ω(A D ).
Proposition 7.16.For all excited diagrams First, we show that for the Young diagram of µ both statistics a(•) and ω(•) agree.
Proof.We proceed by induction on |µ| with λ fixed.When µ = ∅ we have both Now, either directly or by Remark 7.5 for µ = ∅, Let ν be obtained from µ by adding a cell at position (a, b).Then Next, the array A ν is obtained from A µ by moving the ones in diagonal Since h(i, j) = λ i + λ j − i − j + 1, then h(i, j) − h(i, j + 1) cancels λ i − i + 1 and h(i, j) − h(i + 1, j) cancels the terms λ j − j.So by doing horizontal and vertical cancellations on diagonals d k and d k in (7.7) (see Figure 11, Left) we conclude that either if the diagonals d b and d b have the same length, or Then by induction it follows that ω(A ν ) = a(D ν ).
Lemma 7.18.Let D ∈ E(λ/µ) be obtained from D ∈ E(λ/µ) with one excited move.Then a(D Proof.Suppose D is obtained from D by replacing (i, j) by (i + 1, j + 1) then We illustrate these differences in Figure 11: Right.

Excited diagrams and SSYT of border strips and thick strips
In the next two sections we focus on the case of the thick strip δ n+2k /δ n where δ n denotes the staircase shape (n − 1, n − 2, . . ., 2, 1).We first study the excited diagrams E(δ n+2k /δ n ) and the number of SYT of this shape via the NHLF (Theorem 1.2) and our first q-analogue for SSYT of this shape (Theorem 1.4).8.1.Excited diagrams and Catalan numbers.Let FanDyck(k, n) be the set of tuples (γ 1 , . . ., γ k ) of k noncrossing Dyck paths from (0, 0) to (2n, 0) (see Figure 13: Right).We call such tuples k-fans of Dyck paths.We show that excited diagrams in E(δ n+2k /δ n ) are in correspondence with non-crossing Dyck paths.
Proposition 8.1.The number of excited diagrams in E(δ n+2k /δ n ) is equal to the number of fans of Dyck paths in FanDyck(k, n): Proof.We start with the case k = 1.By Proposition 3.7, excited diagrams in E(δ n+2 /δ n ) are in bijection with flagged tableaux of shape δ n with flag (2, 3, . . ., n).It is well known and easy to see that these are in bijection with Dyck paths (row i of these tableaux have entries i and i + 1, the boundary between these values outlines the Dyck path γ), as illustrated in Figure 12.For general k, by the same argument, the excited diagrams in E(δ n+2k /δ n ) are in bijection with flagged tableaux of shape δ n with flag (k + 1, k + 2, . . ., k + n − 1).These tableaux correspond to k-tuples (γ 1 , . . ., γ k ) of k noncrossing Dyck paths .
Proof.By Proposition 3.7, we have On the other hand, the fans of paths in FanDyck(k, n) are counted by the given determinant of Catalan numbers, and also by the given product formula [SV].
From here we easily obtain the following curious determinantal identity (see also §11.7).
Remark 8.4.Note that by Proposition 3.7, excited diagrams in E(δ n+2k+1 /δ n ) are in correspondence with flagged tableaux of shape δ n with flag (k + 1, k + 2, . . ., k In what follows the formulas for the even case δ n+2k are simpler than those of the odd case so we omit the latter. Remark 8.5.Fans of Dyck paths in FanDyck(k, n) are equinumerous with k-triangulations of an (n + 2k)-gon [Jon] (see also [S6, A12] and [SS] for a bijection for general k).
Taking the limit q → 1 in Corollary 8.9 we get corresponding identities for f δ n+2k /δn .
In a different direction, one can use Corollary 8.10 when n = 1, 2 to obtain the following determinant formulas for Euler numbers in terms of for f δ 2k+1 and f δ 2k , which of course can be computed by a HLF (cf.[OEIS,A005118]).
Corollary 8.13.We have: , where E n = E n /n! and (a, b) ∈ γ denotes a point of the Dyck path γ.
Proof.For the LHS we use Corollary 8.10 to express f δ n+2k /δn in terms of Euler numbers.For the RHS, we first use the NHLF to write f δ n+2k /δn as a sum over excited diagrams E(δ n+2k n ) : , where D = [δ n+2k /δ n ] \ D. By Proposition 8.1, excited diagrams of δ n+2k /δ n correspond to k-tuples of noncrossing Dyck paths in FanDyck(k, n).Finally, one can check (see Figure 13) that if D → (γ 1 , . . ., γ k ) then which gives the desired RHS.Let BT (n) be the set of plane full binary trees τ with 2n + 1 vertices, i.e. plane binary trees where every vertex is a leaf or has two descendants.These trees are counted by |BT (n)| = C n (see e.g.[S6, §2]).Given a vertex v in a tree τ ∈ BT (n), h(v) denotes the number of descendants of v (including itself).An increasing labelling of τ is a labelling ω(•) of the vertices of τ with {1, 2, . . ., 2n + 1} such that if u is a descendant of v then ω(v) ≤ ω(u).By abuse of notation, let f τ is the number of increasing labelings of τ .By the HLF for trees (see e.g.[Sag3]), we have: .
Proposition 8.14.We have: Proof.The RHS of (8.6) gives the probability E 2n+1 /(2n + 1)! that a permutation w ∈ S 2n+1 is alternating.We use the representation of a permutation w as an increasing binary tree T (w) with 2n + 1 vertices (see e.g.[S3, §1.5]).It is well-known that w is an down-up permutation (equinumerous with up-down/alternating permutations) if and only if T (w) is an increasing full binary tree [S3,Prop. 1.5.3].See Figure 14 for an example.We conclude that the probability p that an increasing binary tree is a full binary tree is given by p = E 2n+1 /(2n + 1)!.
On the other hand, we have: where f τ /(2n + 1)! is the probability that a labelling of a full binary tree τ is increasing.By (8.5), the result follows.On the other hand, for the two Dyck paths in Dyck(2), Equation (1.11) gives 8.5.q-analogue of Euler numbers via SSYT.We use our first q-analogue of NHLF (Theorem 1.4) to obtain identities for s δ n+2k /δn (1, q, q 2 , . ..) in terms of Dyck paths.
Proof of Corollary 1.10.By Theorem 1.4 for the skew shape δ n+2 /δ n and (8.2) we have Let D in E(δ n+2 /δ n ) corresponds to the Dyck path γ and cell (i, j) in D corresponds to point (a, b) in γ then h(i, j) = 2b + 1 and λ j − i = b.Using this correspondence, the LHS of (8.7) becomes the LHS of the desired expression.
Corollary 8.17. where Proof.For the LHS, use Corollary 8.9 to express s δ n+2k /δn (1, q, q 2 , . ..) in terms of q-Euler polynomials E m (q).For the RHS, first use Theorem 1.4 for the skew shape δ n+2k /δ n and then follow the same argument as that of Corollary 8.13.
Remark 8.18.Let us emphasize that the only known proof of Corollary 8.17 that we have, uses both the algebraic proof of Theorem 1.4, properties of the Hillman-Grassl bijection (see Theorem 7.7), and the Lascoux-Pragacz theorem (Theorem 8.7).As such, this is the most technical result of the paper.

Pleasant diagrams and RPP of border strips and thick strips
In this section we study pleasant diagrams in P(δ n+2 /δ n ) and our second q-analogue of NHLF (Theorem 6.3) for RPP of shape δ n+2 /δ n .9.1.Pleasant diagrams and Schröder numbers.Let s n be the n-th Schröder number [OEIS, A001003] which counts lattice paths from (0, 0) to (2n, 0) with steps (1, 1), (1, −1), and (2, 0) that never go below the x-axis and no steps (2, 0) on the x-axis.Theorem 9.1.have: |P(δ n+2 /δ n )| = 2 n+2 s n , for all n ≥ 1.The proof of Theorem 9.1 is based on the following corollary which is in turn a direct application of Lemma 6.15.Recall that high peak of a Dyck path γ is a peak of height strictly greater than one.We denote by HP(γ) the set of high peaks of γ, and by N P(γ) the points of the path that are not high peaks.We use 2 S denote the set of subsets of S.
Proof.By Lemma 6.15, we have: By the proof of Proposition 8.1 excited diagrams D of shape δ n+2 /δ n are in correspondence with Dyck paths γ in Dyck(n).Under this correspondence D → γ, excited peaks Λ(D) are identified with high peaks HP(γ) and the rest [δ n+2 ] \ (D ∪ Λ(D)) is identified with N P(γ), defined as set of points in γ that are not high peaks.
Proof of Theorem 9.1.It is known (see [Deu]), that the number of Dyck paths of size n with k − 1 high peaks equals the Narayana number N (n, k) = 1 n n k n k−1 .On the other hand, Schröder numbers s n can be written as (9.1) (see e.g.[Sul]).By Lemma 9.2, we have: where we use the symmetry N (n, k) = N (n, n − k + 1) and (9.1).
In the same way as |E(δ n+2k /δ n )| is given by a determinant of Catalan numbers, the preliminary computations suggests that |P(δ n+2k /δ n )| is given by a determinant of Schröder numbers.
Conjecture 9.3.We have: |P(δ n+4 /δ n )| = 2 2n+5 (s n s n+2 − s 2 n+1 ).More generally, for all k ≥ 1, we have: , where s n = 2 n+2 s n .Here we use s n = |P(δ n+2 /δ n )| in place of s n in the determinant to make the formula more elegant.In fact, the power of 2 can be factored out.9.2.q-analogue of Euler numbers via RPP.We use our second q-analogue of the NHLF (Theorem 1.5) and Lemma 9.2 to obtain identities for the generating function of RPP of shape δ n+2 /δ n in terms of Dyck paths.Recall the definition of E * n (q) from the introduction: where κ = (13254 . ..).Note that maj(σκ) is the sum of the descents of σ ∈ S n not involving both 2i + 1 and 2i.
Finally, preliminary computations suggest the following analogue of Corollary 8.9.
Conjecture 9.5.We have: π∈RPP(δ n+2k /δn) , 10.Other formulas for the number of standard Young tableaux In this section we give a quick review of several competing formulas for computing f λ/µ .10.1.The Jacobi-Trudi identity.This classical formula (see e.g.[S3, §7.16]), allowing an efficient computation of these numbers.It generalizes to all Schur functions and thus gives a natural q-analogue for SSYT.On the negative side, this formula is not positive, nor does it give a q-analogue for RPP.
10.2.The Littlewood-Richardson coefficients.Equally celebrated is the positive (subtractionfree) formula where c λ µ,ν are the Littlewood-Richardson (LR-) coefficients.This formula has a natural q-analogue for SSYT, but not for RPP.When LR-coefficients are defined appropriately, this q-analogue does have a bijective proof by a combination of the Hillman-Grassl bijection and the jeu-de-taquin map; we omit the details (cf.[Whi]).
On the negative side, the LR-coefficients are notoriously hard to compute both theoretically and practically (see [Nara]), which makes this formula difficult to use in many applications.10.3.The Okounkov-Olshanski formula.The following curious formula is of somewhat different nature.It is also positive, which might not be immediately obvious.
Denote by RT(µ, ) the set of reverse semistandard tableaux T of shape µ, which are arrays of positive integers of shape µ, weakly decreasing in the rows and strictly decreasing in the columns, and with entries between 1 and .The Okounkov-Olshanski formula (OOF) given in [OO] states: where c(u) = j −i is the content of u = (i, j).The conditions on tableaux T imply that the numerators here non-negative.
It is illustrative to compare the NHLF and the OOF for the shape λ/(1) since f λ/(1) = f λ .The excited diagrams E(λ/(1)) consist of single boxes of the diagonal d 0 of λ, thus the NHLF gives On the other hand, the reverse tableau RT((1), (λ)) are of the form T = i for 1 ≤ i ≤ (λ).For each of these tableaux T we have λ T (1,1) = λ i and c(1, 1) = 0, thus the (OOF) gives Note that in both cases i h(i, i) = i λ i = |λ|, confirming that f λ/(1) = f λ , however the summands involved in both formulas are different in number and kind.Chen and Stanley [CS] found a SSYT q-analogue of the (OOF).Their proof is algebraic; they also give a bijective proof for shapes λ/(1).It would be very interesting to find a bijective proof of the formula and its q-analogue in full generality.Note that again, there is no RPP q-analogue in this case.
On the positive side, the sizes |RT(µ, )| are easy to compute as the number of bounded SSYT of the (rectangle) complement shape µ; we omit the details.10.4.Formulas from rules for equivariant Schubert structure constants.In this section we sketch how there is a formula for f λ/µ for every rule of equivariant Schubert structure constants, a generalization of the Littlewood-Richardson coefficients.
Now, the excited diagrams that appear in the NHLF come from the rule to compute C λ µ,λ = [X w ] v in Theorem 4.3.Moreover, any rule to compute C λ µ,ν (•) gives a formula for f λ/µ .Below we outline two such rules: the Knutson-Tao puzzle rule [KT] and the Thomas-Yong jeu-de-taquin rule [TY].Given partitions λ, µ, ν ⊆ d × (n − d) with |λ| ≥ |µ| + |ν| we consider a tilling of the triangle with edges labelled by the binary representation of the subsets corresponding to ν, µ, λ in [n]   d (clockwise starting from the left edge).To each equivariant piece in a puzzle we associate coordinates (i, j) coming from the coordinates on the horizontal edge of the triangle form SW and SE lines coming from the piece: We denote the piece with its coordinates by p ij .The weight wt(P ) of a puzzle P is wt(P ) = pij ∈P ; eq.
(y i − y j ), where the product is over equivariant pieces.Let ν ∆ µ λ be the set of puzzles of a triangle boundary ν, µ, λ (clockwise starting from the left edge of the triangle) .Knutson and Tao [KT] showed that C λ µ,ν is the weighted sum of puzzles in ν ∆ µ λ .
Note an important disadvantage of (TYF) when compared to LR-coefficients and other formulas: the set of tableaux EqSYT(λ, µ) does not have an easy description.In fact, it would be interesting to see if it can be presented as the number of integer points in some polytope, a result which famously holds in all other cases.
Example 10.8.Consider the case when λ/µ = (2 2 /1).There are two tableaux of shape λ/µ that rectify to the superstandard tableaux 1 2 3 4 of weight λ: 1 2 3 4 and 2 1 3 4 , where the first tableau has weight (2 − 1 + 1 − (0)) = 2 corresponding to edge label 1, and the second tableau with weight (2 − 2 + 1 − (−1)) = 2 corresponding to edge label 3.By Corollary 10.7, we have Comparing with the terms from the NHLF, we have 2 excited diagrams which contributes a weight 3 (hook length of the blue square) and which contributes weight 1, so As this example illustrates, the Thomas-Young formula (TYF) and the NHLF have different terms, and thus neither equivalent nor easily comparable.10.5.The Naruse hook-length formula.In lieu of a summary, the NHLF has both SSYT and RPP q-analogue, both of which have a bijective proof. 3It has a combinatorial proof (via the RPP q-analogue and combinatorics of excited and pleasant diagrams), but no direct bijective proof.It is also a summation over a set E(λ/µ) which is easy to compute (Corollary 3.8).As a bonus it has common generalization with Stanley and Gansner's trace formulas (see §1.5).More curiously, it gives explicit combinatorial formulas for the thick strip cases (see §1.6 and §11.9).

Final remarks
11.1.There is a very large literature on the number of SYT of both straight and skew shapes.We refer to a recent comprehensive survey [AR] of this fruitful subject.Similarly, there is a large literature on enumeration of plane partitions, both using bijective and algebraic arguments.We refer to an interesting historical overview [K4] which begins with MacMahon's theorem and ends with recent work on ASMs and perfect matchings.
There is an even greater literature on alternating permutations, Euler numbers, Dyck paths, Catalan and Schöder numbers, which are some of the classical combinatorial objects and sequences.We refer to [S5] for the survey on the first two, to [S6] for a thorough treatment of the last three, and to [GJ,OEIS,S3] for various generalizations, background and further references.
Finally, the first q-analogue E n (q) of Euler numbers we consider is standard in the literature and satisfies a number of natural properties, including a q-version of equation (1.10) (see e.g.[GJ,§4.2]).However, the second q-analogue E * n (q) appears to be new.It would be interesting to see how it fits with the existing literature of multivariate Euler polynomials and statistics on alternating permutations.
3 To be precise, only the RPP q-analogue (1.6) is proved fully bijectively.We do not have a description of the (restricted) inverse map Ω = Φ −1 to give a fully bijective proof of (1.4).Instead we prove that the (restricted) Hillman-Grassl map is bijective in this case in part via an algebraic argument.We believe that map Ω can in fact be given an explicit description, but perhaps the resulting bijective proof would be more involved (cf.[NPS]).
11.2.As we mention in the introduction, there are many proofs of the HLF, some of which give rise to generalizations and pave interesting connections to other areas (see e.g.[Ban,CKP,GNW,Han,K1,NPS,Pak,Rem,Ver]).Unfortunately, none of them easily adapt to skew shapes.Ideally, one would want to give a NPS-style bijective proof of the NHLF (Theorem 1.2), but for now any direct proof would be of interest.
In a different direction, the hook-length formula for f λ has a celebrated probabilistic proof [GNW].If an NPS-style proof is too much to hope for, perhaps a GNW-style proof of the NHLF would be more natural and as a bonus would give a simple way to sample from SYT(λ/µ) (as would the NPS-style proof, cf.[Sag2]).Such algorithm would be theoretical and computational interest.Note that for general posets P on n elements, there is a O(n 3 log n) time MCMC algorithm for perfect sampling of linear extensions of P [Hub].
11.3.The excited diagrams were introduced independently in [IN1] by Kre2] by Kreiman in the context of equivariant cohomology theory of Schubert varieties (see also [GK,IN2]).For skew shapes coming from vexillary permutations, they also appear in terms of pipe dreams or rc-graphs in the work of Knutson,Miller and Yong [KMY,§5], who used these objects to give formulas for double Schubert polynomials of such permutations.11.4.While a direct bijective proof of NHLF would be the most interesting, even some special cases would be of interest.For example, the proof of Corollary 1.9 we give in this paper is combinatorial but very technical and involves the Hillman-Grassl bijection, Stanley's P-partition theory and the connection between excited and pleasant diagrams.Perhaps, there is a simple proof?The reader may want to compare it to deceptively similar but much simpler Proposition 8.14.11.5.As we mention in the previous section, RPP typically do not arise in the context of symmetric functions.A notable exception is the recent work by Lam and Pylyavskyy [LamP], who defined a symmetric function g λ/µ (x) in terms of RPP of shape λ/µ, and have a LR-rule [Gal].However, these functions are not homogeneous and the specialization g λ/µ (1, q, q 2 , . ..) is different than our RPP q-analogue.11.6.By Corollary 3.8, the number of excited diagrams of λ/µ can be computed with a determinant of binomials.Thus |E(λ/µ)| can be computed in polynomial time.This raises a question whether |P(λ/µ)| can be computed efficiently (see Section 6.1).For example, Theorem 9.1 and Conjecture 9.3 claim that for the border strips and thick strips these numbers can be computed efficiently.Perhaps, Theorem 6.14 can be applied in the general case.11.7.The curious Catalan determinant in Corollary 8.3 is both similar and related4 to another Catalan determinant in [AL,proof of Lemma 1.1].In fact, both determinants are special cases of more general counting results, and both can be proved by the the Lindström-Gessel-Viennot lemma.
For example, in the border strip case, this conforms with the known asymptotics of Euler numbers: (see e.g.[S5] and reference therein).In the opposite extreme, when µ fixed and λ grows, the Okounkov-Olshanski formula works better; Stanley gave several precise asymptotic bounds in this case [S4].
11.9.The connection between alternating permutations and symmetric functions of border strips goes back to Foulkes [Fou], and has been repeatedly generalized and explored ever since (see [S5]).It is perhaps surprising that Corollary 1.9 is so simple, since the other two positive formulas in Section 10 become quite involved.For the LR-coefficients, let partition ν 2n + 1 be such that ν 1 , (ν) ≤ n + 1.
It is easy to see that in this case the corresponding LR-coefficient is nonzero: c δn+2 δn ν > 0, suggesting that summation over all such ν would can be hard to compute.
Similarly, the Okounkov-Olshanki formula appears to have a large number of terms with no obvious pattern.It seems, just like in the asymptotic considerations above, (OOF) is best when |µ| is small, while the NHLF is best when |λ/µ| is small, compared to |λ|.11.10.Along with Theorem 1.2, Naruse also announced a formula for the number g λ/µ of standard tableaux of skew shifted shape λ/µ, in terms of type D excited diagrams.These excited diagrams are obtained from the diagram of µ by applying the following two excited moves and the NHLF then extends verbatim.
, .It would be of interest to find both q-analogues of this formula, as in theorems 1.4 and 1.5.Let us mention that while some arguments translate to the shifted case without difficulty (see e.g.[K1,Sag1]), in other cases this is a major challenge (see e.g.[Fis]).
Definition 3.5 (Interlacing property).Let D ⊂ [λ].If (i, j) and (i + m, j + m) are two consecutive elements in D ∩ d k then D contains an element in each diagonal d k−1 and d k+1 between columns j and j + m.Note that the excited diagrams in E(λ/µ) satisfy this property by construction.

4.
Algebraic proof of Theorem 1.4 4.1.Preliminary results.A skew shape λ/µ with µ ⊆ λ ⊆ d × (n − d) is in correspondence with a pair of Grassmannian permutations w v of n both with descent at position d and where is the strong Bruhat order.Recall that a permutation v = v 1 v 2 • • • v n is Grassmannian if it has a unique descent.The permutation v is obtained from the diagram λ by writing the numbers 1, . . ., n along the unit segments of the boundary of λ starting at the bottom left corner and ending at the top right of the enclosing d × (n − d) rectangle.The permutation v is obtained by first reading the d numbers onthe vertical segments and then the (n − d) numbers on the horizontal segments.The permutation w is obtained by the same procedure on partition µ (see Figure3).

Figure 5 .
Figure 5. Two consecutive cells a and b in S that violate the interlacing property of excited diagrams.

Figure 6 .
Figure 6.Example of the maps 1 and 2 on a pleasant diagram S.

Figure 8 .
Figure 8.The diagonals d 1 (µ), . . ., d (λ)−1 (µ), the support of A µ represented by diagonals, and the support of array A D associated to an excited diagram D.
7.2.Φ −1 (A * D ) are RPP of skew shape.Given an excited diagram D ∈ E(λ/µ), let A D be the set of arrays of nonnegative integers of shape λ with support in [λ] \ D. Note that the set of excited arrays A * D from Definition 7.6 is contained in A D .We show that the RPP in Φ −1 (A D ) have support contained in λ/µ and therefore so do the RPP in Φ −1 (A * D ).Lemma 7.10.For each excited diagram D ∈ E(λ/µ), the reverse plane partitions in Φ −1 (A * D ) have support contained in λ/µ.Proof.By Lemma 6.7, the support of arrays in A D are pleasant diagrams in P(λ/µ).So by Theorem 6.3 it follows that Φ −1 (A D ) ⊆ RPP(λ/µ).Since A * D ⊆ A D , the result follows.7.3.Φ −1 (A * D ) are column strict skew RPP.Lemma 7.11.For each excited diagram D ∈ E(λ/µ), the reverse plane partitions in Φ −1 (A * D ) are column strict skew RPPs of shape λ/µ.

Figure 11 .
Figure 11.The equality of statistics a(D) and ω(A D ).

Figure 14 .
Figure 14.The full binary tree corresponding to the alternating permutation w = (6273514).